1

### JEE Main 2019 (Online) 9th January Morning Slot

Let
A = $\left\{ {\theta \in \left( { - {\pi \over 2},\pi } \right):{{3 + 2i\sin \theta } \over {1 - 2i\sin \theta }}is\,purely\,imaginary} \right\}$
. Then the sum of the elements in A is :
A
${5\pi \over 6}$
B
$\pi$
C
${3\pi \over 4}$
D
${{2\pi } \over 3}$

## Explanation

Given complex number,

${{3 + 2i\sin \theta } \over {1 - 2i\sin \theta }}$

$= {{\left( {3 + 2i\sin \theta } \right)\left( {1 + 2i\sin \theta } \right)} \over {1 + 4{{\sin }^2}\theta }}$

$= {{3 + 6i\sin \theta + 2i\sin \theta - 4{{\sin }^2}\theta } \over {1 + 4{{\sin }^2}\theta }}$

$= {{\left( {3 - 4{{\sin }^2}\theta } \right) + i\left( {8\sin \theta } \right)} \over {1 + 4{{\sin }^2}\theta }}$

As complex number is purely imaginary, So real part of this complex number is zero.

$\therefore$   ${{3 - 4{{\sin }^2}\theta } \over {1 + 4{{\sin }^2}\theta }}$ = 0

$\Rightarrow$   $3 - 4{\sin ^2}\theta = 0$

$\Rightarrow$   $\sin \theta = \pm {{\sqrt 3 } \over 2}$

as   $\theta$ $\in$ $\left( { - {\pi \over 2},\pi } \right)$

$\therefore$   $\theta$ $=$ $-$ ${\pi \over 3},{\pi \over 3},{{2\pi } \over 3}$

$\therefore$   Sum of those values of A is

$= - {\pi \over 3} + {\pi \over 3} + {{2\pi } \over 3}$

$= {{2\pi } \over 3}$
2

### JEE Main 2019 (Online) 9th January Evening Slot

Let z0 be a root of the quadratic equation, x2 + x + 1 = 0, If z = 3 + 6iz$_0^{81}$ $-$ 3iz$_0^{93}$, then arg z is equal to :
A
${\pi \over 4}$
B
${\pi \over 6}$
C
${\pi \over 3}$
D
0

## Explanation

1 + x + x2 = 0

x = ${{ - 1 \pm \sqrt {1 - 4} } \over 2} = {{ - 1 \pm i\sqrt 3 } \over 2}$

z0 = w, w2

Now

z = 3 + 6iz$_0^{81}$ $-$ 3iz$_0^{93}$

z = 3 + 6iw81 $-$ 3iw93      (w93 = w81 = 1)

$\Rightarrow$  z = 3 + 3i

then arg(z) = tan$-$1$\left( {{3 \over 3}} \right)$ = tan$-$1 (1) = ${\pi \over 4}$
3

### JEE Main 2019 (Online) 10th January Morning Slot

Let z1 and z2 be any two non-zero complex numbers such that   $3\left| {{z_1}} \right| = 4\left| {{z_2}} \right|.$  If  $z = {{3{z_1}} \over {2{z_2}}} + {{2{z_2}} \over {3{z_1}}}$  then -
A
${\rm I}m\left( z \right) = 0$
B
$\left| z \right| = \sqrt {{17 \over 2}}$
C
$\left| z \right| =$ ${1 \over 2}\sqrt {9 + 16{{\cos }^2}\theta }$
D
Re(z) $=$ 0

## Explanation

Given, $3\left| {{z_1}} \right| = 4\left| {{z_2}} \right|$

$\Rightarrow$ ${{\left| {{z_1}} \right|} \over {\left| {{z_2}} \right|}} = {4 \over 3}$

$\Rightarrow$ ${{\left| {3{z_1}} \right|} \over {\left| {2{z_2}} \right|}} = {4 \over 3} \times {3 \over 2} = 2$

As we know, for any compled number

${{3{z_1}} \over {2{z_2}}} = {{\left| {3{z_1}} \right|} \over {\left| {2{z_2}} \right|}}$(cos$\theta$ + i sin$\theta$)

= 2(cos$\theta$ + i sin$\theta$)

$\therefore$ ${{2{z_2}} \over {3{z_1}}}$ = ${1 \over {2\left( {\cos \theta + i\sin \theta } \right)}}$

= ${1 \over {2\left( {\cos \theta + i\sin \theta } \right)}} \times {{\left( {\cos \theta - i\sin \theta } \right)} \over {\left( {\cos \theta - i\sin \theta } \right)}}$

= ${\left( {{1 \over 2}\cos \theta - {i \over 2}\sin \theta } \right)}$
Now, given

$z = {{3{z_1}} \over {2{z_2}}} + {{2{z_2}} \over {3{z_1}}}$

= 2(cos$\theta$ + i sin$\theta$) + ${\left( {{1 \over 2}\cos \theta - {i \over 2}\sin \theta } \right)}$

= ${{5 \over 2}\cos \theta + {3 \over 2}i\sin \theta }$

So, |z| = $\sqrt {{{25} \over 4}{{\cos }^2}\theta + {9 \over 4}{{\sin }^2}\theta }$

= ${1 \over 2}\sqrt {9 + 16{{\cos }^2}\theta }$

z is neither purely real nor purely imaginary and |z| depends on $\theta$.
4

### JEE Main 2019 (Online) 10th January Evening Slot

Let $z = {\left( {{{\sqrt 3 } \over 2} + {i \over 2}} \right)^5} + {\left( {{{\sqrt 3 } \over 2} - {i \over 2}} \right)^5}.$ If R(z) and 1(z) respectively denote the real and imaginary parts of z, then -
A
R(z) = $-$ 3
B
R(z) < 0 and I(z) > 0
C
I(z) = 0
D
R(z) > 0 and I(z) > 0

## Explanation

$z = {\left( {{{\sqrt 3 + i} \over 2}} \right)^5} + {\left( {{{\sqrt 3 - i} \over 2}} \right)^5}$

$z = {\left( {{e^{i\pi /6}}} \right)^5} + {\left( {{e^{ - i\pi /6}}} \right)^5}$

$= {e^{i5\pi /6}} + {e^{ - i5\pi /6}}$

$= \cos {{5\pi } \over 6} + i{{\sin 5\pi } \over 6} + \cos \left( {{{ - 5\pi } \over 6}} \right) + i\sin \left( {{{ - 5\pi } \over 6}} \right)$

$= 2\cos {{5\pi } \over 6} < 0$

${\rm I}(z) = 0$ and ${\mathop{\rm Re}\nolimits} (z) < 0$